HDU 2665 Kth number

Problem Description

Give you a sequence and ask you the kth big number of a inteval.

 

Input

The first line is the number of the test cases. 
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere. 
The second line contains n integers, describe the sequence. 
Each of following m lines contains three integers s, t, k. 
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

 

Output

For each test case, output m lines. Each line contains the kth big number.

 

Sample Input

1 10 1 1 4 2 3 5 6 7 8 9 0 1 3 2 

 

Sample Output

2
主席树的应用，也称函数式线段树
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=100005;int n,m,T,a[maxn],b[maxn],c[maxn];int L[maxn*20],R[maxn*20],sum[maxn*20],tot,frist[maxn];bool cmp(const int &x,const int &y){    return a[x]<a[y];}void insert(int now,int l,int r,int u){    sum[++tot]=sum[now]+1;    if (l==r) L[tot]=R[tot]=0;    else     {        int mid=(l+r)>>1;        L[tot]=L[now];    R[tot]=R[now];        if (u<=mid) L[tot]=tot+1; else R[tot]=tot+1;        if (u<=mid) insert(L[now],l,mid,u);        else insert(R[now],mid+1,r,u);    }}int query(int u,int v,int l,int r,int k){    if (l==r) return a[b[l]];    else     {        int mid=(l+r)>>1;        if (sum[L[u]]-sum[L[v]]<k)             return query(R[u],R[v],mid+1,r,k-(sum[L[u]]-sum[L[v]]));        else             return query(L[u],L[v],l,mid,k);    }}int main(){    scanf("%d",&T);    while (T--)    {        scanf("%d%d",&n,&m);        for (int i=1;i<=n;i++) scanf("%d",&a[b[i]=i]);        sort(b+1,b+n+1,cmp);        for (int i=1;i<=n;i++) c[b[i]]=i;        memset(frist,0,sizeof(frist));        frist[0]=L[0]=R[0]=sum[0]=tot=0;        for (int i=1;i<=n;i++)         {            frist[i]=tot+1;            insert(frist[i-1],1,n,c[i]);        }        while (m--)        {            int x,y,k;            scanf("%d%d%d",&x,&y,&k);            printf("%d\n",query(frist[y],frist[x-1],1,n,k));        }    }    return 0;}